The following got no answer on mathstackexchange. I believe it not to be hard, but maybe it is a little specialized?

All varieties will be over $\mathbb{C}$ and projective unless stated otherwise.

In Beauville - complex algebraic surfaces, the following is described: Let $S$ be a smooth surface and $p \in S$ a point. Let $\epsilon: \tilde S \rightarrow S$ be the blowup at $p$ and $E$ the resulting exceptional curve. Then $$ \text{Pic}(\tilde S) \cong \epsilon^* \text{Pic}(S) \oplus \mathbb{Z} E $$ With $\text{Pic}$ i mean either the group of invertible sheaves or those of Cartier divisors modulo equivalence.

**Question 1:** I was wondering about the situation when $p$ is a simple double singularity, a node, instead of being smooth and the rest of $S$ is smooth. Does the same formula hold? If not, is there is a similar formula that describes $\text{Pic}(S)$ as a direct summand of $\text{Pic}(\tilde S)$?

**Question 2:** Does anybody know a reference for this situation: relationship of Picard groups of singular surfaces (or varieties in general) with their smoothification?

**My guess and thoughts so far:**

My first guess was that the same would hold, but i think that is false. The question is treated locally in Hartshorne example 6.5.2, which examines $$ \text{Spec}(\mathbb{C}[x,y,z]/(xy - z^2)) $$ The Weil class group of this affine variety is $\mathbb{Z}/2\mathbb{Z}$ and is generated by a ruling of the cone. This ruling is not a Cartier divisor and the cartier divisor class group is trivial.

This makes me conjecture: $$ \text{WCl}(\tilde S) \cong \epsilon^* \text{Wcl}(S) \oplus \mathbb{Z} E $$ $$ \text{Pic}(\tilde S) \cong \epsilon^* \text{Pic}(S) \oplus \mathbb{Z} R \oplus \mathbb{Z} E $$ where $R$ corresponds to a ruling of the cone, which was a weil divisor on the singular variety but after the blowup corresponds to a Cartier divisor as well. $\text{WCl}$ means the group of Weil divsors modulo equivalence.

This is just an intuitive guess coming from Hartshorne's example so please please correct me if i'm wrong.

Thanks!